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B.Tech. IInd YEAR

IVth Semester




This is to certify that Mr./Ms./Mrs._______________________________________________ Reg. No./Roll No. ______________________of ___________________________________________ class has satisfactory completed the course in _________________________________ at College of Agricultural Engineering & Technology, Anand Agricultural University, Godhra.

Date of Submission: ____________________________________

Concern Faculty: ________________________________________

Head of Department: __________________________


Name Symbol Quantity Expression in terms of other


Expression in terms of SI base

units Newton N Force, Weight m·kg/s2 m·kg.s-2

Joule J Energy, work, heat N·m m2·kg.s-2 Watt W Power J/s m2·kg.s-3

Pascal Pa Pressure, stress N/m2 m-1·kg.s-2

Some Important Derived Units Name Symbol Quantity Expression in terms of SI

base units Square meter m2 Area m2 Cubic meter m3 Volume m3

Meter per second m.s-1 Speed, velocity m.s-1 Meter per second

square m.s-2 Acceleration m.s-2

Radian per second

rad.s-1 Angular velocity rad.s-1

Newton second N.s Momentum impulse


Newton meter second

N.m.s Angular momentum


Newton meter N.m Torque, moment of force


Kilogram per cubic meter

kg.m-3 Density, mass density


Cubic meter per kilogram

Kg-1.m3 Specific volume Kg-1.m3

Mass moment of inertia

kg.m2 Mass moment of inertia





Name of Experiment Date of



1 Introduction to Theory of machine & its Basic terms

2 Introduction to cam & cam profile & to solve example of cam profile

3 To study about velocity & acceleration & to draw velocity & acceleration

diagram for given data.

4 Introduction of friction & its Applications.

5 Int. of power Transmission & to solve examples on power Transmission

6 Study of turning moment dia. Fly wheel & to solve the example of flywheel.

7 To Study of Governor.

8 Study of Balancing




Importance of theory of machine in design and analysis. Fundamental of theory of machine Basic mechanism

INTRODUCTION:-The subject of theory of machine may be defined as that branch of engineering which

deals with the study of relative motion between the various parts of a machine and forceswhich act on them.

Thus to study the motion of different parts of a mechanisms, the study of forces is notnecessary and can be neglected. The study of mechanisms can be divided in to thefollowing disciplines.

Theory of machine

Kinematics Dynamics

Statics KineticsKinematics:-

It deals with the relative motions of different parts of a mechanism without takinginto consideration the forces producing the motions. Applications are the displacement,velocity and acceleration of a part of a mechanism.Dynamics:-

It is that branch of theory of machines which deals with the forces and there effectwhile acting upon the machine parts in motion.Kinetics:-

It is that branch of theory of machines which deals with the inertia forces whicharise from the combined effect of the mass and the motion of the machine parts.Statics :-

It is that branch of theory of machines which deals with the forces and their effectwhile the machine parts are at rest. The mass of the parts is assumed to be negligible.

►VARIOUS TERMSKinematic link

A mechanism is made of a number of resistant bodies out of which some havemotion relative to the others. A resistant body or a group of resistant bodies with rigidconnection preventing there relative movement is known as a link.

►TYPES OF LINKS1) Rigid link:-A rigid link is one which does not undergo any deformation while transmitting motion.Strictly speaking, rigid link does not exist. Connecting rod, crank e.t.c are rigid links.



2) Flexible link:-A flexible link is one which is partly deformed in a manner not to effect the transmissionof motion. Belts, ropes, chains, e.t.c are flexible links.

3) Fluid links:-A fluid link is one which is formed by having a fluid in a receptacle and the motion istransmitted through the fluid by pressure or compression only, as in case of hydraulicpresses, jacks and brakes.


Kinematics pairs can be classified according to,Nature of contactNature of mechanical constraintNature of relative motion

According to nature of contact

Lower pairA pair of links having surfaced or area of contact between the members known as

a lower pair. The contact surfaces of the two links are similar. E.g. -nut turning on ascrew, shaft rotating in a bearing, all pairs of a slider crank mechanism, universal joint,e.t.c.

Higher pairWhen a pair has a joint or line contact between the links is known as a higher

pair. the contact surfaces of the two links are dissimilar. e.g.-wheel rolling on a surface,cam and follower pair, tooth gears ball and roller bearings etc.

According to nature of mechanical constraint

Closed pairWhen the elements of a pair are held together mechanically, it is known as a

closed pair. The two elements are geometrically identical, one is solid and full and theother is hollow or open. The latter not only envelopes the former but also encloses it. Thecontact between the two can be broken only by destruction of at east one of the members.

All the lower pairs and some of the higher pairs are closed pairs. Cam andfollower pair and screw pair are closed pair.

Unclosed pairWhen two links of a pair are in contact either due to force of gravity or some spring

action, constitute an unclose pair. In this, the links are not held together mechanically.E.g. cam and follower pair.




According to nature of relative motion

Sliding pairIf two links have a sliding motion relative to each other, the form a sliding pair.

A rectangular rod in a rectangular hole in a prism is a sliding pair.

Turning pairWhen one link has a turning or revolving motion relative to the other, they constitute aturning or revolving pair.In a slider crank mechanisms, all pair expect the slider and guide pair are turning pairs. Acircular shaft revolving inside a bearing is a turning pair.

Rolling pairWhen the links of a pair have a rolling motion relative to each other, they form a

rolling pair. E.g. a rolling wheel on a flat surface, ball and roller, bearings etc..In a ballbearing the ball and the shaft constitute one rolling pair where as the ball and a bearing isthe second rolling pair.

Screw pairIf two matting links have a turning as well ad sliding motion between them, they

from a screw pair. This is a achieved by cutting machining threads on the twolinks.e.g.lead screw and nut of a lathe is a screw pair.

Spherical pairWhen one link in the form of a sphere turns inside a fixed link, it is a spherical pair.

The ball and socket joint is a spherical in fig.

►Kinematic chainWhen the kinematic pairs are coupled in such a way that the last link is joined to the

first link to transmit definite motion, it is called as kinematic chain.

►MechanismWhen one of the links of a kinematic chain is fixed, the chain is known as

mechanism. It is used for transmitting motion. E.g. typewriterA mechanism with four links is known as simple mechanism and the mechanism

with more than four links is known as compound mechanism.

►MachineWhen a mechanism is required to transmit power or to do some partic ular type of

work, if then becomes machines. A machine transforms the available energy into someuseful work.


►Questions :-

1. Define the following terms by giving examples.(a) Elements of link (b) Lower pair & Higher pair(c) Kinematic chain (d) Mechanism(e) Link (f) Quadric chain(g) Kinematic Pair. (h) Kinematics & Kinetics.(i) Theory of machine.

2. Enlist the inversion of double slider crank chain. Explain any one with neatSketch.

3 Define Machine. State how it differs from mechanism.

4. Explain completely constrained motion by giving example.

5. Explain Single slider crank mechanism and its any one in version.

6. Difference between structure & Machine

7 Difference between Machine & Mechanism . What is meant by inversion ofmechanism?

8. Difference between Mechanism & Inversion.






Introduction to am profile. Terminology of cam profile Cam and profile.

►INTRODUCTIONA cam is a mechanical member used to impart desired motion to a follower by

direct contact. The cam may be rotating, reciprocating or oscillating. Complicated outputmotions which are otherwise difficult to achieve can easily be produced wit h the help ofcams. cams are widely used tools, printing control mechanisms and so they aremanufactured usually by die casting, milling or by punch presses.

A cam and the follower combination belong to the category of higher pairs.Necessary elements of a cam mechanism are,(A)Driver member known as the cam.(B)Driven member called the follower.(C)Frame which supports the cam and guides the follower.

►Types of camsWedge camsFlat camsRadial or disc camsCylindrical or barrel or drum cam

►Types of followerCam follower are classified according to theShapeMovementLocation of line movement

●According to shape▪Knife edge follower▪Roller follower▪Mushroom follower▪Flat face follower

●According to movement▪Reciprocating follower▪Oscillating follower

●According to location of line movement▪Radial follower▪Offset follower




►Base circle0It is the smallest circle tangent to the cam profile drawn from the centre of rotation of aradial cam.

►Trace pointIt is reference point on the follower to trace the cam profile such as the knife edgefollower and centre of the roller of a roller follower.

►Pitch circleIt is the circle passing through the pitch point and concentric with base circle

►Pitch curveIt is curve drawn by the trace point assuming that the cam is fired and the trace point ofthe follower rotates around the cam.

►Pitch pointIt is the point on the pitch curve at which the pressure angle is found maximum.

►Pressure angleThe pressure angle, representing the steepness of the cam profile is the angle between thenormal to the pitch curve at appoint and th e direction of the follower motion. It varies inmagnitude at all instant of the follower motion. A high value of the maximum pressureangle is not desired as it might jam the follower in the bearing.

►Prime circleThe smallest circle drawn tangent to the pitch curve is known as the prime circle.


1. Explain with neat sketches different types of cams & Followers.2. Explain with neat sketch how a cam operates a valve in IC Engine.3. What is the function of cam? What are the components of cam mechanism.4. Why different types of follower motion are used.


Problem :-1Draw the profile of cam required for operating, exhaust value of an oil engine. It isrequired to give S.H.M. during the operating & closing of the value each of which iscorresponds to 60o of cam rotation. The value must remain in fully open position for 20 o

of cam rotation. The lift of the value is 36 mm & the list value of radius of cam is 50 mm.The follower is provided with a roller of 40 mm diameter and its line of stroke is passesto the axis of cam. Assume the clockwise rotation of cam.

Problem :-2Draw the cam profile to move the knife edge follower to give a lilt of 40 mm with S.H.M.during the 90o of cam rotation. Then it returns with uniform acceleration and retardationduring 120o of cam rotation. The follower remaining in rest for remaining period. Theaxis of follower passes to the axis of cam shaft. The cam rotates in clockwise direction.The basic circle diameter of cam is 60 mm.

Problem :-3Draw the cam profile with following data for knife edge follower which is moving alonga redial line.

(1) Outward stroke take place with SHM during 120 o of cam rotation.(2) The follower remain at rest in the highest position during 30 o of cam rotation.(3) Return stroke take place during 90 o of cam rotation with SHM and follower

remain at rest for the remaining period.The stroke of follower is 50 mm and base circle radius 40 mm. The cam rotates inclockwise direction with uniform speed.

Problem :-4Draw the profile of cam that give a lilt of 40 mm to a rod carrying a 20 mm dia. roller.The axis of the roller passes to the centre of the cam. The least radius of cam is 50 mm.The rod is lifted with SHM in 1/3 th of cam rotation, dwells for 1/12 th of rotation. &descents with uniform velocity in 1/4 th revolution. The cam rotates in clockwisedirection.

Problem :-5From last university exam paper -2009.




Learning and understanding

Understand velocity and acceleration problem

Construct velocity and acceleration diagram.

►Linear velocity:

It may be defined as the rate of change of linear displacement of a body with respect tothe time. Since velocity is always expressed in a particular direction therefore it’s avector quantity.


Linear velocity, V = ds / dt

►Linear acceleration:-

It may be defined as the rate of change of change of linear velocity of a body with respectto the time


Linear acceleration , a = dv / dt

►Angular Velocity:-

It may be defined as the rate of change of angular displacement with respect to time. It’susually expressed by a Greek letter (omega).Since it has magnitude and direction therefor It’s a quantity. It may be represented by a vector following the same rule as describein the previous article.


Angular velocity, = dθ/ dt


►Angular Acceleration;-

It may be define as the rate of change of angular velocity with respect to time. It’s usuallyexpressed by a Greek letter (Alpha). Since it has magnitude and direction both thereforeit’s a quantity. It may be represented by a vector following the same rule as describe inthe previous article.


Angular Acceleration, = d / dt

►Relation between Linear & Angular Quantity of Motion: -

Consider a body moving along a circular path from A to B as shown .

Let, r = Radius of the circular path

= Angular displacement in radian.

S = Linear displacement

V = Linear velocity

= Angular velocity.

A = Linear acceleration

= Angular acceleration


Motion of a body along a circular path

From the geometry of the figure.

We know that, s = r

We also know that the linear velocity

V=ds/dt = d (r) / dt =r.d / dt

V = .r

Linear Acceleration

a = dv/dt = d (.r) / dt = r.d / dt

a = r.



[1] Four Bar Chain Mechanism

Draw the space diagram on scale as per given measurement

-- AD is a fixed link

-- AB is a Driver (crank)

-- Angle θ is given

-- Direction of rotation is also given

Draw the velocity diagram

-- Select suitable scale

-- ω = 2πN / 60 rad / sec. where N is in r.p.m.

-- VB = ω AB m/sec ( AB is measurement of link AB in meter)

-- A & D are fixed point so, a & d are same



-- Draw VB perpendicular to link AB from point a & equal to VB on some scale

i.e ab.

-- From b, draw a line perpendicular to link BC

-- From a & d draw a line perpendicular to link CD

-- Both line intersect each other at c

-- Linear velocity of link BC = VCB = bc * velocity scale

-- Linear velocity of link CD = VC = cd * velocity scale

-- Angular velocity of link BC = ω.CB = VCB / BC

-- Angular velocity of link CD = ω.C = VC / CD

-- Velocity of point E on link BC (BE is given)


BE / BC = be / bc

so, be =______mm

-- Plot point e on bc

-- Joint a and e measure ae

-- VE = ae * scale mm/sec.

Draw the Acceleration diagram

-- Link AB has uniform rotational motion, so there is no tangential components

of angular acceleration.

-- Radial component of link AB = a2BA = a B = V2

BA./ BA = V2.

B / BA m/sec2.

( VB from velocity triangle m/sec & BA length of link AB in meter)

-- Radial component of link BC = arCB = V2

CB / CB m/sec2.

-- Radial component of link CD = ar CD = V2C / CD m/sec2


-- Select a suitable scale for acceleration diagram

-- Take any point a’& d’

-- From a’ draw a B & parallel to AB

-- From b’ draws b’x equal to arCB & parallel to CB

-- From x draw a line perpendicular to b’x ___________ _____[1]

-- From d’draw d’y equal to arCD & parallel to CD

-- From y draw a perpendicular to d ’y ___________________[2]

-- Line (1) & (2) Intersect on c’

-- Joint c’ and a’ and c’ and b’

-- atCB = x c’ * scale m/sec2

-- atCD = y c’ * scale m/sec2


-- Angular acceleration of link BC = α BC = atCB / CB

-- Angular acceleration of link CD = α CD = atCD / CD

-- Acceleration of point of a E on link BC :-

BE / BC = b ’e’ / b’ c’

So, b’ e’ = ___________mm.

-- Joint e’ and a’

-- Acceleration of E = a E = a’ e’ * scale.



[2] Slider crank Mechanism

Draw the space diagram on scale as per given measurement

-- Crank radius (r) OB

-- Angular position (θ) from I.D.C

-- Length of connecting rod AB (C.R.)

ω = 2πN / 60 rad/sec

= angular velocity, where N is in r.p.m

-- Direction of rotation

-- Select suitable scale

Draw the velocity diagram

-- VB = ω. r = ω * OB m/sec

-- Select suitable scale


-- Take any point o. draw VB from o & perpendicular to OB & get b

-- From b draw a line perpendicular to link AB

-- From o draw a line parallel to the path of slider i.e. parallel to OA

-- Both lines intersects at a.

-- Velocity of slider (Piston) A = VA

= oa * scale

-- Velocity of link AB = VBA

= ba * scale

-- Angular velocity of link AB = ωAB

= VBA / AB

-- Velocity point Eon link AB

be / ba = BE / BA

So be = _______m/sec

-- Plot point e on ba and join with o

-- Velocity of point E = oe * scale.

Draw the Acceleration diagram

-- The crank OB has uniform rotation motion so there are no tangential

component of motion.

-- Radial components of acceleration = a2BO = a B = V2

BO / BO = V2B / BO

-- Select a suitable scale

-- Take any point o’. Draw o ’b’ parallel to BO & equal to a B on a scale

-- From b’, draw b’x’ parallel to AB & equ al to

arAB = V2


-- From x’ draw a line perpendicular to b’x’_________________[1]


-- From o’ draw a line parallel to AO which intersect [1] in point a’

-- Acceleration of piston A = a A = o ’a’ * scale.

-- Acceleration of connecting rod AB = a AB = b ’a’ * scale

-- Acceleration of point E on C. R.

a’ e’ / a’ b’ = AE / AB

b’e’ / b’ a’ = BE / BA

-- Plot e’ and join with o’

-- Acceleration of point E = o’ e’ * scale.

-- Angular acceleSration of C.R. AB = α AB = atAB / AB = (a ’x’ * scale) / AB

Klein construction for velocity and acceleration of the points ofReciprocating Engine


-- OB crank radius

-- AB connecting rod length

-- θ Angle

-- RPM-N of angular velocity at ω rad/sec.

-- Direction of rotation (clockwise/anticlockwise)

-- Position of point E

(Use some scale to the path of piston)


-- Extend AB beyond B and draw perpendicular from o which intersects in M.

-- OBM is a Klein’s velocity diagram

-- Take B as a centre and BM as a radius and draw the circl e.

-- Draw another circle taking AB is a diameter, both circle intersects at K & L



-- Join K & L and extend it which intersect the path of piston i.e. line AO in N .

KL intersect AB in Q.

-- Quadrilateral BQNO is known as Klei n’s acceleration diagram

Velocity calculation :- ( OBM)

-- Velocity of crank = V OB = ω * OB * scale

-- Velocity of piston = V OA = ω * OM * scale

-- Velocity of C.R. = VAB = ω * BM * scale

-- Velocity of Pin E on C.R is given by,

BE / AB = BE1 / BM BE = ________

-- Join E, and O.

-- Velocity of point E = VE = ω * OE1 * scale

Acceleration calculation :- ( Quadrilateral BQNO )

-- Radial acceleration of crank = arBO

= ω2 * OB * scale

-- Acceleration of Piston = a OA

= ω2 * NO * scale

-- Radial acceleration of C.R = a rAB

= ω2 * BQ * scale

-- Tangential acceleration of C.R. = a tAB

= ω2 * QN * scale

-- Total Acceleration of C.R. = a AB

= ω2 * BN * scale


-- For acceleration of point E on C.R.

Draw a line from E parallel to the path of piston

i.e. AO, which intersect BN into E 2

-- Join O & E2

-- Acceleration of point E = a E = ω2 * OE2 * scale.


1. Write short note on relative velocity method.

2. In velocity & acceleration diagram state the relation between linear & angularquantities such as displacement , velocity & acceleration.

Problem :-1

In four chain ABCD, AD is fixed link 120 mm long, driving link AB is 30 mm long androtates uniformly at 100 r.p.m clockwise, while the link CD is 60 mm long & oscillatesabout D. Coupler link BC is equal as link AD. Draw velocity diagram & find the angularvelocity of link DC, when angle BAD is 60 o.

Problem :-2

The crank & connecting rod of steam engine are 0.5 m & 2.0 m respectively. The crankrotates at 200 rpm. In clockwise direction. While crank has turned 60 from IDC , Findgraphically using relative velocity method, the following,

(1) Velocity & acceleration of piston.

(2) Angular velocity & angular acceleration of connecting rod

(3) Velocity & acceleration of amid point of connecting rod.


Solve the above problem using Klei n construction method.


Problem :-3

The crank & connecting rod of reciprocating Engine are 150 mm & 600 mm respectively.The crank in rotating in clockwise direction at 120 rad / sec. Crank position in at 45 o

from IDC. Find out graphically (1) Velocity & acceleration of the piston. (2) Velocity &acceleration of the mid point of the connecting rod.


Solve the above problem using Klein construction method.

Problem :-4

In the simple steam engine the length of crank & connecting rod are 100 mm & 400 mmrespectively. The centre of mass of connecting rod is 220 mm from the crank pin center.If the engine speed is 400 rpm. Determine graphically absolute velocity & acceleration ofthe C.G. of connecting rod when the crank has turned 45 o clockwise from IDC.


Solve the above problem using Klein construction method.

Problem :-5

From last university exam paper – 2009.





Learning and understanding. Basic concept of friction Motions of the body in different plane. Friction and working different types of clutches, brakes, bearings Dynamometer

{ A } Definition & motion on inclined plane

►Introduction :-No surface is perfectly smooth. Even most accurately machined surface will also

have ridges & depression. When one surface comes in direct contact with anothersurface., their ridges & depression are interlocked and therefore a ny effort to move onesurface over the other is resisted. This resisting force which opposes the motion is knownas friction.

Friction always resists the motion, therefore some power is lost in friction duringpower transmission.

“ In general the friction is defined as resistance encountered when body slides ortends to slide over the surface of another body. ”►TYPES OF FRICTION1)Static friction:-It is the friction, experienced by a body, when body is at rest.2)Dynamic friction:-It is the friction experienced by a body when body is in motion. The dynamic friction isalso called Kinetic friction and is less than the static friction.Dynamic friction is of the followi ng three types

A) Sliding friction :-It is the friction experienced by a body when it slides over another body.

B) Rolling friction :-It is the friction experienced between the surfaces which have balls or rollersinterposed between them.

C) Pivot friction :-It is the friction experienced by a body due to the motion of rotation as in case ofFoot step Bearings.

z►Limiting Friction :-Consider that a body A of weight W is l ying on a rough horizontal body B as shown infig (a). In this position, body A is in equilibrium under the action of its own weight W,and the normal reaction Rn of B on A. Now a small horizontal force P1 is applied to thebody, acting through its center of gravity as shown in fig (b). It does not move because ofthe frictional force which prevents the motion. This shows that the applied force is lessthan friction F1.



If we increase the applied force P1 to P2 as shown in fig (c), it is found to be inequilibrium. This means that the force of friction has also increased to a value F2 = P2.Thus every time the effort is increased , the force of friction is also increases, so as tobecome exactly to be applied force.

After this, any increased in applied effort , the body A begins to move in the direction ofapplied force. This maximum value of friction force, when a body just begin slide overthe surface of other body, is known as “Limiting Force of Friction.” as shown in fig (d).It may be noted that when the applied force is less than the limiting friction, the bodyremains at rest, and the friction is called Static Friction which may have any valuebetween zero and Limiting Friction.

►Co-efficient of friction:It is the defined as the ratio of the limiting friction (F) to the normal reaction (Rn)between the two bodies. It is generally denoted by µ .Mathematically:

Co-efficient of friction µ = F / Rn

►Limiting angle of friction: -The resultant reaction R ,equal and opposite to the resultant of W & P and will beinclined at an angle Φ to the normal reaction Rn. This angle Φ is known as the limitingangle of friction. It may be defined as the angle which the resultant reaction R makeswith the normal reaction Rn.

Tan Ø = F / Rn = µ Rn / Rn = µ

►Angle of repose:-Consider that a body A of weight (W) is resting on an inclined plane B, as shown in fig.If the angle of inclination α of the plane to the horizontal is such that the body begins tomove down the plane, then the angle α is called the angle of repose.A little consideration will show that the body will begin to move down the plane whenthe angle of repose is equal to the angle of friction (α = Ø). This may be proved asfollows.The weight of the body(W) can be resolved into the following to components:

1. Wsinα, parallel to the plane B. this components tends to slide the body down the plane.

2. Wcosα, perpendicular to the plane B.this component is balanced by the normalreaction (Rn) of the body A and the plane B , ( Wcosα = Rn ).

When, Wsinα = F = µRn = µWcosα

tan α = µ = tan Ø or α = Ø



►Laws of static friction:-1) The force of friction always acts in a direction opposite to that in which the body

tends to move.2) The magnitude of the force of friction is exactly equal to the force, which tends

the body to move.3) The magnitude of the limiting friction (F) bears a constant ratio to normal reaction

(Rn) between the two surfaces. Mathematically F / Rn = Constant4) The law of force of friction is independent of the area of the contact between the

two surfaces.5) The force of friction depends upon the roundness of the surface.

►Laws of Kinetic or Dynamic Friction1) The force of friction always acts in a direction, opposite to that in which the body

is moving.2) The magnitude of kinetic friction bears a constant ratio to the normal reaction

between the two surfaces. But th is ratio slightly less than in case of Limitingfriction.

3) For moderate speeds, the force of friction remain constant. But it decreasesslightly with increase in speed.

►Motion of Body on a Horizontal Plan e.

Minimum force required to slide a body ona rough horizontal surface is given by,

P = Wsin Ø / Cos(Ө - Ø)

►Motion of Body up on inclined planBy neglecting friction

According to the Sine rule,

Po / Sinα = W / Sin(Ө - α )

Po = W Sinα / Sin(Ө - α ) _______(1)

By considering frictionP = W Sin (α + Ø) / Sin ( Ө - (α + Ø)) __(2)

Where,Po = Force effort by neglecting friction.P = Force effort by considering frictionα = Angle of ResponseӨ = Angle between Weight & Line of Action.Ø = Angle between Resultant & Normal Reaction.

Rn = Normal ReactionR = Resultanance Reaction


Motion of Body up in inclined plane With effort P & Po Actingparallel to Slop

Motion of Body up in inclined plane with effort P & Po Actingin horizontal Direction


►Motion of Body up in inclined plane With effort P & Po Acting parallel to Slop

By neglecting friction,

Ө = ( 90 + α )Put the value of Ө in above eq.- (1), so we get,

Po = W Sinα / Sin90 = W Sinα _____(3)

By considering friction,

Ө = ( 90 + α )Put the value of Ө in above eq.- (2), so we get,

P = W Sin( α + Ø ) / CosØ ______(4)

►Motion of Body up in inclined plane with effort P & Po Acting in HorigentalDirection

By neglecting friction,

Ө = 90

Put the value of Ө in above eq.- (1), so we get,

Po = W tanα ______(5)

By considering friction,

Ө = 90

Put the value of Ө in above eq.- (2), so we get,

P = W tan (α + Ø) ____(6)

►Motion of Body Down on inclined plane

By neglecting frictionAccording to the Sine rule,

Po / Sinα = W / Sin(Ө - α )

Po = W Sinα / Sin(Ө - α )( Same as eq-(1) )


Motion of Body Down on inclined plane

Motion of Body Down on inclined plane With effort P & PoActing parallel to Slop


By considering friction

According to Lami’s RulesP = W Sin (Ø - α) / Sin ( Ө + (Ø - α)) ____(7)

►Motion of Body Down on inclined plane With effort P & Po Acting parallel toSlop

By neglecting friction,

Ө = ( 90 + α )Put the value of Ө in above eq.- (1), so we get,

Po = W Sinα / Sin90 = W Sinα

By considering friction,

Ө = ( 90 + α )Put the value of Ө in above eq.- (7), so we get,

P = W Sin(Ø - α ) / CosØ

►Motion of Body Down on inclined plane with effort P & Po Acting in HorigentalDirection

By neglecting friction,

Ө = 90

Put the value of Ө in above eq.- (1), so we get,

Po = W tanα

By considering friction,

Ө = 90

Put the value of Ө in above eq.- (7), so we get,

P = W tan (Ø - α ) ____(8)

►Efficiency of Inclined Planeη = Ratio of force effort by neglecting friction to the force effort by considering

friction.= Po / P


Types of Thread

{ B } Screw Jack & Thrust Bearing

►Helix :-It is the curve trace by a point in circular way with uniform veloc ity and magnitude in

axial direction is called Helix.Ex : screw thread.

►Pitch :-The distance between two nearby similar point of the thread in axial direction is know asa pitch.

►Lead :-The distance traveling in one revolution by nut in axial di rection is known as a lead.

►Multi Threaded Screw : -It is the thread in which more than one thread are cut in one lead distance.In Single thread screw Lead = Pitch.In Double threaded screw, two thread are cut in one Lead.In Triple threaded screw, three thread are cut in one Lead.

S = p * ZTanα = p / π d

Where, S = Leadp = Pitch

Z = No. of thread in lead distance ( No. of Start )α = Helix angled = Mean dia. of screw

►Types of Screw thread:-1) V- Thread2) Square thread3) Trapezoidal or Acme thread4) Buttress thread


Screw Jack :-1) Find out the effort P required to lift up the load in horizontal direction.2) Find out the effort P required to lower down the load in horizontal direction .3) Find out the Torque T required to lift up the load by Screw Jack.4) Find out the Torque T required to lower down the load by Screw Jack .



Thrust Bearings




Thrust Bearing :-In case of rotating shaft , there may be a force acting along the axis of shaft which isknown as “thrust”. The thrust produce lateral motion of shaft along the axis , which canbe prevented by one or more bearing surfaces at right angles to the axis of rotation. Thesesurfaces are known as thrust bearings. The bearing which carry the axial thrust are knownas thrust bearings.►Classification :-

Thrust Bearing

Pivot Bearing Collar Bearing

Flat Pivot Conical Pivot Truncated Pivot

Single Flat Collar Multi Flat Collar►Equations :-Friction Torque required for different Thrust Bearing is as given below,


Types of Bearing Uniform PressureP = constant

Uniform WearP * r = constant

1 Truncated conicalThrust Bearing T =

32 µ Sin


T =

21 µ

SinW {R1+R2}

2 Conical pivotThrust Bearing(R2 =0 )

T =32 µ

SinWR T =

21 µ



3 Collar ThrustBearing(α = 90 )

T =32 µ Sin


T =32 µ W {R1+ R2 }

4 Foot Step / FlatPivot Bearing(α = 90 , R2 = 0 )

T =32 µ W R T =

32 µ W R

Where,W = Total Axial LoadR = Radius of Bearing Surface

R1 = Outer radius of CollarR2 = Inner radius of Collar

Power lost in Friction for different Thurst Bearing is given by,

P = 2 π N T / 60000 in KW.


Single Plate Clutch Cone Clutch

{ C } Clutches , Brakes , & Dynamometer.

Clutches:-A Clutch is a device used to transmi t the rotary motion of one shaft to another whendesired. The axis of the two shaft are coinciding.In Clutches, the connection of Engine shaft to the gear b ox shaft is affected by thefriction between two or more rotating concentric surfaces. The surfaces are pressedfirmly against one another when engaged and the clutch tend to rotate as a single unit.

►Classification :-Clutch

Mechanical Clutch Electromagnetic Clutch Hydraulic Clutch

Positive Contact Clutch Friction Clutch

Disc or Plate Clutch Cone Clutch Centrifugal Clutch

Single Plate Clutch Multi Plate Clutch

►Friction Torque and power transmitted in Single and multi plate clutch is givenby,For Uniform pressure

W = p π ( R12 – R22 )T = 2/3 * µ W {R13 – R23 / R12 – R22}* NeT = µ W Rm NeP = 2 π N T / 60000 in KW.


For Uniform WearW = 2π p R2 ( R1 – R2 )T = 1/2 * µ W {R1+ R2 }* NeT = µ W Rm NeP = 2 π N T / 60000 in KW.

Where,R1 = External Radius of Friction plate in mmR1 = Internal Radius of Friction plate in mmNe = Number of active ( contact ) surfaceW = Total axial load in NP = Intensity of pressure in N/mm2

T = Friction torque in N. m►Friction Torque and power transmitted in Cone clutch is given by,For Uniform pressureW = p π ( R12 – R22 )T = 2/3 * µ W / Sinα {R13 – R23 / R12 – R22 }P = 2 π N T / 60000 in KW.

For Uniform WearW = 2π pmax R2 ( R1 – R2 )W = 2π pmax R2* b SinαT = 1/2 * µ W / Sinα {R1+ R2 }T = µ W / Sinα RmP = 2 π N T / 60000 in KW.

Where,R1 = External Radius of Friction plate in mmR1 = Internal Radius of Friction plate in mmNe = Number of active ( contact ) surfaceW = Total axial load in NP = Intensity of pressure in N/mm2

T = Friction torque in N. mα = Semi cone angle

Rm = (R1 + R2) / 2►Brakes :-Brake is a machine element or a devi ce through which artificial frictional resistant isapplied to the body in motion, as a result the moving body comes to rest or its speed isreduced. Brake absorbs the Kinetic Energy.

Classification :- Brakes

Pneumatic Brake Hydraulic Brake Electric Brake Mechanical Brake

Band Brake Block / Shoe Brake Band & Block Brake Cone Brake Disc Brake

External Block / Shoe Brake Internal Block / Shoe Brake

Simple Band Brake Differential Band Brake


Simple Band BrakeBand Brake

►Friction Force and Friction Torque in Simple Band Brake is given by,Tension Ratio = T1/T2 = e µӨ

Friction Force/ Braking Force = T1- T2

Friction Torque / Braking Torque = Braking Force * Radius of DrumMt = ( T1 – T2 ) * R

Applied Force ( P ) , when drum rotates in clockwise direction. ( a = 0 )P * L= T2 * b

P = ( T2 * b )/ LApplied Force ( P ) , when drum rotates in anticlockwise direction. ( b = 0 )

P * L= T1 * a

P = ( T1 * a )/ LWhere,

T1 = Tension in tight side in NT2 = Tension in slack side in NӨ = Angle of LapR = Radius of drum

►Friction Force and Friction Torque in Differential Band Brake is given by,Tension Ratio = T1/T2 = e µӨ

Friction Force/ Braking Force = T1- T2

Friction Torque / Braking Torque = Braking Force * Radius of DrumMt = ( T1 – T2 ) * R


Differential Band Brake

Applied Force ( P ) , when drum rotates in clockwise direction. ( a = 0 )

P * L + T1 * a = T2 * b

P = ( T2 b – T1 a )/ L

P = T2{ b - (T1/T2)*a} / L

P = T2{ b - e µӨ * a} / L

when b/a = e µӨ then P = 0 and brake become self lockingwhen b/a < e µӨ then P becomes -Ve i.e. brake called self energizing brake

Applied Force ( P ) , when drum rotates in anticlockwise direction. ( b = 0 )

P * L + T2 * b = T1 * a

P = ( T1 a – T2 b )/ L

P = T2{ (T1/T2)*a - b} / L

P = T2{ e µӨ* a - b} / L

when b/a = e µӨ then P = 0 and brake become self lockingwhen b/a < e µӨ then P becomes -Ve i.e. brake called self energizing brake


Band and Block Brake

Prove thatTn / To = { 1 + µ tanӨ / 1 - µ tanӨ }n for Band and Block brake






DynamometerA Dynamometer is a brake incorporating devise to measure the frictional resistanceapplied. Dynamometer is used to determine the power developed by machine or primemover, while maintaining its speed at rated value.


Absorption type Transmission type

Mechanical Hydraulic Mechanical Electrical

Prony Brake Rope Brake Belt Transmit ion Epicyclic gear train

Prony Brake DynamometerTorque Mt = F * R

Moment of Resistance = W * L

Mt = F * R = W * L

Work done in one revolution = Tork * Angle in Radian= Mt * 2π

Work done in one minute = Mt * 2π N

Bp = Break Power = Work done in one minute / 60 = 2πNMt / 60

Bp = W L * 2π N / 60 in WattsWhere,

W = Weight at the outer end of the lever in N.L = Horizontal distance of weight W from the center of pulley in meter

R = Radius of pulley in meterN = Speed of shaft in r.p.m.F = Frictional resistance between blo ck & Pulley in N.

Mt = Torque in N-m.

Rope Brake Dynamometer

Brake horse power Bp = {( W- S ) (D + d ) * π N } / 60 in Watts.Where,

W = Dead load in N.S = Spring balance reading in ND = Diameter of Wheel in meterd = Diameter of rope in meter


N = Speed of shaft in r.p.m.

►Questions :-

1. State the law of solid & Dry friction.2. Show in case of truncated conical pivot, for uniform wear condition, the f riction forcemay be assumed to be acting at an effective radius given by

r1 + r2 / 2 sinαwhere α = Semi cone angle

r1 + r2 = Smallest & Largest radius.3. Explain any one type of absorption dynamometer with a neat sketch. State also theequation for power developed by engine.4. State the law of friction. What are the advantage of friction ?5. State the function of clutch and sketch any one type of clutch used in practice.6. Derive an expression for centrifugal tension developed in belt while transmittingpower.7. Explain the operation of band & block brake. Explain also action when a brake workas an dynamometer.8. List the various types of thrust bearing and give their uses.9. Derive the expression of friction torque for a flat pivot bearing with usual notationassuming uniform pressure condition.10. State the law of friction. Define the limiting an gle of friction and write the relationbetween coefficient of friction & limiting angle of friction.11. State the function of clutch. By neat sketch explain the single plate clutch.12. Explain the construction & working of rope brake dynamometer.

Promlem :-1Mean diameter of single start square threaded screw jack is 60 mm. The pitch of thread is10 mm. The coefficient of friction is 0.15. What force must be applied at the end of 0.9 mlong lever to raise a load of 20 KN and to lower it.

Promlem :-2A multi plate clutch of alternate bronze and steel plat having effective diameter 175 mm& 72.5 mm has to transmit 22 KW at 2000 rpm. The end thrust is 1700 N and coefficientof friction is 0.1. Calculate the number of plate necessary, assuming uniform pres suredistribution on the plate.

Promlem :-3A differential band brake has drum diameter 350 mm. The ends of the band are fixed tothe pin on opposite side of the fulcrum of a lever at a distance of 35 mm & 150 mm fromthe fulcrum. The angle of contact is 225o. The brake has to sustain a torque of 350 N -m.and coefficient of friction between band and CI drum is 0.30. Find the necessary force forclockwise & anticlockwise rotation of drum which is to be applied at the lever at distanceof 500 mm from fulcrum.


Problem :- 4A thrust shaft of a ship has a 6 collar of 600 mm external diameter and 300 mm internaldiameter. The total thrust from propeller is 120 KN. If coefficient of friction is 0.12 andspeed of engine is 1000 rpm. Find the power absorbed in f riction at the thrust blockassuming (i) uniform pressure (ii) uniform wear condition. What condition is morefavorable? Why?.

Problem :- 5A single start square thread screw jack is used to lift the load. The load of 220 N isapplied to at the end of its lever 500 mm long. Calculate the load which can lifted. If it isplaced on the head of swivel of 100 mm dia. The load is not rotating with spindle. Theroot dia of screw & pitch of screw are 50 mm & 10 mm respectively. The frictioncoefficient between screw & nut is 0.18. The friction coefficient between spindle &swivel head is 0.15. Find out the efficiency of screw jack.

Problem :- 6The mean dia of square thread screw jack is 40 mm. The pitch of thread is 8 mm. Thecoefficient of friction between screw & nut is 0.18. and between platform & thrustbearing is 0.15. A load of 20 KN is carried on a platform having bearing mean dia 70mm. Calculate (i) Total torque required to raise the load. (ii) Force to be applied at theend of 0.7 m long lever. (iii) Ef ficiency of machine.

Problem :- 7A shaft of a ship rests on a multi collar thrust bearing having 500 mm external dia & 250mm internal dia. The total thrust on a bearing is 90 KN. If the coefficient of frictionbetween shaft & bearing is 0.12 and speed o f engine is 120 rpm. Calculate the power lostin friction by considering (i) uniform pressure condition. (ii) uniform wear condition.




Learning and understanding Introduction, need, modes and applications. Belt drive, gear train.

►Introduction :The belt or ropes are used to transmit power from one shaft to another by means ofpulleys which rotate at the same speeds. The amount of power transmitted depends uponthe following factors.

-- The velocity under which the belt is placed on the pulleys.-- The tension under which the belt and the sm aller pulley.-- The arc of contacts between the belt and the smaller pulley.-- The condition under which the belt is used it may be noted that.

(a) The shaft should be properly in line to insure uniform tension across the beltsection.

(b) The pulley should not be too close together ,in order that the arc of contact onthe smaller pulleys may be as large as possible

(c) The pulleys should not be so far apart to cause the belt to weigh heavily on theshaft, thus increasing the friction load on the bearings.

(d) A long belt tends to swing from side to side to side, causing the belt to run outof the pulleys. Which is run develops croocke spots in the belt.

(e) The tight side of the belt should be at the bottom, so that whatever say ispresent on the loose side will increase the arc of contact at the pulleys.

(f) In order to obtain good results with flat belts, the maximum distance betweenthe shafts should not exceed 10 meters and the minimum should not be lessthan 3.5 times the diameter of the larger pulley.

►Types of Belt drivesThe belt drives are usually classified in to the following three groups.

1. Light drives: - These are used to transmit small powers at belt speeds up to about10 m/s as in agricultures machines and small machines tools.

2. Medium drives: - These are used to transmit medium power at belt speeds over 10m/s but up to 22 m/s in machines tools.

3. Heavy drives: - These are used to transmit large power at belt speeds above 22m/s, as in the compressors and generator.

►Types of Belts & Drives

[1] Belt Drive:-In belt drive Driver and Driven pulley are connected by a end less beltTwo types of belt drive is used

a) Flat Belt: - Flat belt used where, moderate power transmission is required,from one pulley to another. Mostly it is used in factories where the distancebetween to two pulleys is not to be exceeding 8 meters apart


Types of Belt Drive


b) V-Belt:- V-Belt is used where ,moderate power transmission is required, fromone pulleys to another .Mostly it is used in factory workshop, wh ere thedistance between two pulleys is not to be exceeding 2 meters

Advantages of Belt drive:--- Belt drive can be used for two eccentric shafts.-- Belt made from flexible material.-- Less vibration.-- Lubrication is not required.-- It is cheapest drive.Disadvantages of Belt drive:--- Power lost due to slip of belt-- Uniform power can not be obtain-- More space required as compared to other devices.-- Less life.Applications:--- It is used where speed ratio between 5 to 10. And peripheral speed between

10 m /s to 30 m/s.-- Also, It is used where the distance between dr iver and driven pulleys is more

[2] Rope or Circular Drive :- Rope or circular belt is used where great amount ofpower is to be transmitted. Mostly it is used where the distance between two pulleysare more than 8 meters.

[3] Chain Drive: -Chain drive is known as positive drive. It is used where the distancebetween the driver and driven pulleys is short. . In chain drive sprocket is usedinstead of pulleys .chain made from carbon steel .

Advantages of Chain drive:--- Uniform power can be obtain due positive drive.


-- Minimum load on shaft and bearing.-- Less space is required as compared to the belt drive.-- Longer life.-- More power transmission ef ficiency than Belt drive.Disadvantages of Chain drive:--- Costly than belt drive-- More vibration problem than belt drive-- Lubrication is required for this drive.-- Effective length is changed due to friction between chain and sprocket.Application of Chain drive:--- Bicycle, Motorcycle, Printing press/m/c. Textile machine etc

[3] Gear Drive :- Gear drive also known as positive drive. Gear drive is mostly usedwhere power transmission distance from one shaft to another shaft is to less.

Advantages of Gear drive:--- No slip problem-- Uniform power can be obtain-- More efficiency as compared to other drive.-- Gear drive can be used to transmit maximum power.Disadvantages of Gear drive:--- Gear drive can not be used for longer distance pow er transmissionApplication of Gear Drive: --- In machine tool gear-box , Automobile Engine etc.

► Velocity Ratio

-- For open drive:-Both Driver and Driven pulley rotate in same direction

-- For cross belt drive:-Both Driver and Driven pulley rotate in opposite direction

So, Velocity Ratio = πdN1 = πdN2= N1 / N2 = D / d

Where, N1 = Speed of Driver pulleyd = Diameter of Driver pulley

N2 = Speed of Driven PulleyD = Diameter of Driven pulley

By considering Belt thickness,

Velocity ration is given by N 1 / N2 = (D + t) / (d + t)►Velocity ratio of compound belt drived1 = diameter of driver pulley-1, N1 = rpm of driver pulley-1d2 = diameter of driven pulley-2, N2 = rpm of driven pulley-2



d3 = diameter of driver pulley-3, N3 = rpm of driver pulley-3d4 = diameter of driven pulley-4, N4 = rpm of driven pulley-4

Now from fig. pulley 2 & 3 are fitted on a common shaftSo, that N2 = N3.

Now, for stage-1 , N2 / N1 = d1 / d2for stage -2 , N4 / N3 = d3 / d4

So (N2 * N4) / (N1 * N3) = d1 * d3 / d2 * d4But, N2 = N3

N4 / N1 = d1 * d3 / d2 * d4 or N1 / N4 = d2 * d4 / d1 * d3

► Slip :-The forward motion of belt without carrying the driven pulley with it, is called slip ofBelt. It is expressed in %S.

► Method of reducing Slip:-Belt material is flexible so after long time of operation belt stretches and due to thisincreases the length of belt and this effects creates slip. So for reducing slip the centredistance between two pulleys is adjusted according to the following equation,

Le = L - 0.015LBelt dressing should be necessary in belt after the use of long timeProtect the belt against oil & grees.

►Effect of slip on velocity ratio : -S1 = Slip between driver & belt in %.S2 = Slip between driven & belt in %.

S = Total slip = S1+S2.

Velocity ratio = N1 / N2 = d2 / d1 (1-S/100).By considering belt thickness (t) , N1 / N2 = {(d2+t) / (d1+t)} * 1 / (1-S/100)

Power transmitted by belt , P = (T1 -T2) * V / 1000 in kw.where T1 = tight side tension.

T2 = slack side tension.


For open belt drive, θ = (180-2α) * π / 180 rad.where sinα = (r2-r1) / C

L = π/2 * (d1+d2) + 2C + (d2-d1)2 / 4C

For close belt drive, θ = (180+2α) * π / 180 rad.where sinα = (r2 + r1) / C

L = π/2 * (d1+d2) + 2C + (d2+d1)2 / 4C


Tension Ratio of Flat belt Tension Ratio of V belt

Effect of Centrifugal Tension on Belt drive


Centrifugal tension on belt drive (Tc) Tc = m * V2

Where m = Belt mass per meter in Kg / m.V = Belt speed in m / sec.

Max Stress in belt σmax = Tmax / (b * t)Where Tmax = Maximum force in belt (Tension)

b = belt widtht = belt thickness

Belt speed for transmitting maximum power V = √T max / 3 * m

Ratio of driving tension for V-belt T1 / T2 = e µө.cos ß

= 2.3 log (T1/T2) = µ ө.cos ßRatio of driving tension for rope drive, T1 / T2 = e µө.cos ß

= 2.3 log (T1/T2) = µ ө.cos ß

Initial tension in the belt (To) To = {T1 + T2 + 2TC} / 2

Effect of Initial tension at maximum power on belt speed V = √To / 3 * m

►Creep of belt:-When belt passes from slack side to tight side, a certain portion of the belt extends &when belt passes from tight side to slack side, a certain portion of the belt contracts. Dueto this change of length and relative motion between belt and pulley surface occurs. Thisrelative motion is termed as creep of belt.Creep reduces the speed of driven pulley sli ghtly.

By considering creep velocity ratio is given byN2 / N1= {d1/d2 * ( E+ √σ2 / E+ √σ1)}

Where σ1 = Stress in belt on tight sideσ2 = Stress in belt on tight side

E = Young modulus for material of belt

Que-1 for open belt drive, find out angle of contact (θ), length of belt(L)Que-2 for cross belt drive,find out angle of contact (θ),length of belt(L)Que-3 Prove that ratio of tight side tension & slack side tension T1/T2 = eµθ.


Simple Gear Train Compound Gear Train



►Introduction :-Some time two or more gear are made to mesh with each other to transmit power

from one shaft to another such as combination is called “gear train” or train of geartoothed wheels. The nature of train used depends upon the velocity ratio required and therelative position of the axes of the shafts. A gear train may be consist of spur, bevel orspiral gears.

►Types of Gear Train :-Following are the different types of gear trains, depending upon the arrangement ofwheels.

1) Simple gear train2) Compound gear train3) Riveted gear train4) Epicyclic gear train

In first three types of gear train the axis of the shaft over which the gears are mounted arefixed relative to each other . But in case of epicyclic gear train, the axis of the shaft onwhich the gears are mounted may move relative to a fixed axis.

►Simple Gear Train :-When there is only one gear on each shaft, as shown in fig. it is known as simple geartrain. The gears are represented by their pitch circles.When the distance between two shafts is small, the two gear -1 & gear-2 are made tomesh with each other to transmit motion from one shaft to the other, as shown in fig.Since the gear-1 is drive the gear-2 therefore the gear-1 is called driver and gear-2 iscalled driven or follower. It may be noted that the motion of driver gear is opposite to themotion of driving gear.

N1 = Speed of gear-1 in r.p.m.N2 = Speed of gear-2 in r.p.m.T1 = No. of teeth on gear-1.T2 = No. of teeth on gear-2.

Speed Ratio = Speed of driver / Speed of driven

Train Value = Speed of driven / Speed of driver

►Compound Gear Train :-When there is more then one on a shaft, as shown in figure. It is called compound train

of gear.We have seen in that idle gears, in a simple train of gear do not affect the speed ratio of

the system .But these gears useful in dragging over the space between the drivers anddriven.


But when ever the distance between the driver and the driven or follower has to bebridged over by intermediate gears and at the same time a great (or much less) speed rati o

is required, then the advantages of intermediate gears is intensified by providingcompounding gears on intermediate shaft has two gears rigidly fixed to it so the mayhave the same speed. One of these two gears meshes with the driver and with the driv eror follower attached to the next shaft as shown in figure.

Speed ratio = Speed of the first driver/ Speed of the last driven of Follower.

Train value = Speed of the last driver=n or follower/Speed of the first driver.

Thus gear 1 and 2 must have th e same module as they mesh together. Similarly gear 3and 4 and 5 must have the same module.

The speed ratio of compound gear train is obtained by multiplying the equations.

N1 / N2 * N3 / N4 * N5 / N6 = T2 / T1 * T4 / T5 * T6 / T5OR

N1 / N6 = T2 * T4 * T6 / T1 * T3 * T5

►Reverted Gear train:-When the axis of the first gear and the last gear are co -axial, then the gear train is knownas reverted gear train as shown in figure.We see that gear drives the gear 2 in the opposite direction. Since the gear 2 and 3 aremounted on the same shaft, there fore they form a compound gear and the gear 3 willrotate in the same direction as that of gear that of gear 2. The gear 3 drives the gear 4 inthe same direction as that of gear 1. Thus we see that in a reverted gear train the motionof the first gear and the last gear is like.

Let, T1 = Number of teeth on gear 1,R1 = Pitch circle radius of gear 1,N1 = Speed of gear 1 in rpm.

Similarly,T2, T3, T4 = Number of teeth on respective gears.R2, R3, R4 = Pitch circle radii of respective gears andN2, N3, N4 = Speed of respective gears in rpm.Since the distance between the center of the shafts of gears 1 and 2 as well as gear 3 and4 is same, therefore.

R1 + R2 = R3 + R4Also the circular pitch or module of all the gears i s assumed to be same; therefore no. ofteeth on each gear is directly proportional to its circumference or radius.

T1 + T2 = T3 + T4Speed ratio =Product of no. of teeth on driver/product of no. of teeth on driv en.

N1 / N4 = T2 * T4 / T1 * T3



►Epicyclic Gear Train :-We already discuss that in an epicyclic gear train, the axis of the shafts, over which thegear are mounted, may be move relative to a fixed axes. A simple epicyclic gear train isshown in fig. Where a gear A & arm c have a common axis at O1 about which they canrotate. The gear B meshes with gear A and has its axis on the arm at O2, about which thegear B can rotate, of the arm is fixed, the gear train is simple.

Speed Ratio = Speed of first driver / Speed of last drivenSo, NA / ND = 12.

Also NA / ND = NA / NB = NC / NDFor NA / NB & NC / ND to be same , each speed ratio should be √12 so that

NA / ND = NA / NB * NC / ND= √12 * √12= 12

►Questions :-

1. Show that for maximum power transmission the centrifugal tension should notexceed 1/3 rd of total rotation.

2. Define slip. State the effect of slip. State also method of reducing slip.3. What are the advantage & disadvantage of V belt drive ever flat belt drive.4. Explain with neat sketch conical type automobile gear box.5. Derive the expression for centrifugal tension developed in belt while transmitting

Power.6. State the type of gear train & explain any one.7. Derive the expression of belt tension ratio of tension on two side of flat belt in

terms of coefficient of friction & angle of lap.8. List the advantages & limitation of gear drive.9. What is centrifugal tension in belt ? How does it affects the power transmitted.

Problem :-1The following data refer to a flat belt drive :Power transmitted 18 KW , pulley diameter 1.8 m , angle of contact 175 o , speed of pulley300 rpm , coefficient of friction belt & pulley surface is 0.3 0 , permissible stress for belt3 * 106 , thickness of belt 8 mm , density of belt material 0.95 kg / m 3. Determine thewidth of belt required , taking centrifugal tension into account.


Problem :-2A flat belt 8 mm thick & 100 mm wide transmit power between pulleys, running at 25 m/sec. the mass of belt is 0.9 kg per meter length. The angle of lap for smaller pull ey is 165and coefficient of friction is 0.30. The maximum stress for belt material is 2 * 10 6 N / m2.Calculate the maximum power that can b e transmitted.

Problem :- 3The following data refer to a flat belt drive :

-- Maximum pull in belt = 2.0 KN.-- Diameter of driving pulley = 0.5 m.-- Diameter of driven pulley = 0.1 m.-- Speed of driver pulley = 400 rpm.-- Center distance between two shaft = 4.0 m.-- Coefficient of friction = 0.12.-- Mass of belt 1.25 kg/m length.

Find the power transmitted by belt.

Problem :- 450 KW power is required to be transmitted by flat belt pulley of 1.5 m dia and rotating at400 rpm. The angle of contact on pulley is 2.88 radian. The coefficient of frictionbetween pulley & belt is 0.3 m. The thickness of belt is 9.5 mm & mass density of beltmaterial is 1100 Kg / m 3 considering 2.5 N / mm2 as a safe stress and centrifugal tensionin belt. Calculate the width of belt.

Problem :- 5The left handed thread are required to be cut on a job. The pi tch of thread is required0.75 mm. lead screw is having right handed thread of 6 mm pitch. The gear having 20 to120 teeth are available on the lath machine in 5 steps. Find out required gear train.

Problem :- 6Calculate the power transmitted by a V belt having groove angle 38 o and angle of contact165o , the belt speed is 15 m / sec & centrifugal ten sion is 195 N. Maximum tension inbelt is 2200 N. Show your calculation.

Problem :- 7In flat belt drive calculate the power transmitted by belt from following data.

-- Angle of lap = 170o.-- Coefficient of friction between belt & pulley = 0. 25-- Pulley diameter = 100 mm.-- Pulley rpm. = 500-- Maximum tension in belt = 2400 N.





Learning and understanding Moments and moments diagram Flywheel , function , types and moment of inertia. Governor, function , types and termin ology.

►Introduction:The turning moment diagram (also known as crank -effort

diagram) is the graphical representation of the turning, moment of crank effortfor various position of the crank. It is plotted on Cartesian co -ordinates,in which the turning moment is taken as the ordinate and crank angle as abscissa.

►T u r n i n g m o m e n t d i a g r a m f o r a s i n g l e c y l i n d e r D oubleacting steam engine:

The turning moment diagram for a single acting steam engine is shown infigure the vertical ordinate represents the turning momen t and thehorizontal ordinate represents the crank angle

The turning moment on the crank shaft,T = f P X r { S inӨ + S in 2Ө / 2√n 2 – S i n 2 Ө }

Where, fp = Piston effortr = Radius of crankn = Ratio of the connecting rod length and

radius of crankӨ = A n g l e t u r n e d b y t h e c r a n k f r o m i n n e r

d e a d c e n t r eFrom the above expression, we see that turning moment (T) is

zero, when the crank angle ( Ө ) is zero. It is maxim ize when the crank is90° and it is again zero when crank angle is 180°.

►Tu rn i n g m o m e n t d i a g ra m f o r a f o ur s tro k e cy cl e internalcombustion engine:

A t u r n i n g m o m e n t d i a g r a m f o r a f o u r s t r o k e c y c l ei n t e r n a l combustion engine is shown in fig. We know that in a four strokecycle internal combustion engine, there is one working stroke after the


crank has turned through to revolution, i.e. 720°.S i n c e t h e p re s s u r e i n s i d e t h e e n gi n e c yl i n d e r i s l es s t h a n

t h e atmospheric pressure during the suction stroke theref ore a negativelope is obtained. During the expansion or working stroke the fuel burns andthe gases expand, therefore a lar ge positive loop is obtained. In this stroke,the work is d on e b y t h e gas es. Du ri n g ex h au st st ro k e, t h e work i s d on eo n t h e gas es, therefore a negative loop is formed. it may be noted that theeffect of the inertia force on the piston is taken into account in figure.



►Co-efficient fluctuation of energy :It may be defined as the ratio of the max imum fluctuation

of energy to the work done by per cycle.Mathematically,Co-efficient of fluctuation of energy,

CE = Maximum fluctuation of energy work done per cycleThe work done per cycle (in N -m or Joule) may be obtained

by using the following two relations.Work done per cycle = T mean x Ө

where, T mean = Mean torqueӨ = Angle turned in one revolution

= 2 π, In case of steam engine -& two stroke internal combustion engine

= 4 π, In case of four stroke internal combustionEngines.

The mean torque (T mean ) in N -m may be obtain by using followingrelations:T mean = P x 60 / 2 π N= P / W

where. P = Power transmitted in watts.N = Speed in rpm.W = Angular speed in rad/sec

= 2 π N / 60

The work done per cycle may also be obtained by using the following relation.Work done per cycle = P x 60 / n

Where. n = No. of working strokes per minute

The following table shows the value of co efficient of fluctuation of energy forsteam engine and internal combustion engine.


►Flywheel:A flywheel is used in machine serves as a reservoir, which stores e n e r g y d u r i n gt h e p e r i o d w h e n t h e s u p p l y o f e n e r g y i s m o r e t h a n o n e requirement andrelations during the period when the requirement is more than the supply.

►T u r n i n g m o v e m e n t d i a g r a m f o r a m u l t i c y l i n d e r engine:A separate turning moment diagram for a compound steam engine having

three cylinders and the resultant turning moment diagram is shown in fig. T h e r e s u lt a t a n t turning moment diagram is the sum of the turning Moment diagram forthe three cylinders. It may be noted t hat diagram for the three cylinders. It thefirst cylinder is the high pressure cylinder; Second cylinder is the intermediate cylinderand the low pressure cylinder. The cranks, in case of three cylinders are usuallyplaced at 120° to each other.

►Fluctuation of energy:The fluctuation of energy may be determined by the Turning

moment diagram for one complete cycle of operation. Consider the turningmoment diagram for a single cylinder double acting steam engine.

►Co-efficient of fluctuation of energy:It may be defined as the ratio of the max imum fluctuation of

energy to the work done per cycle mathematically.Cc = maximum fluctuation of energy / work done per cycle

►Co-efficient of fluctuation of speed :The difference between the maximum and minim um fluctuation of

speed is caked maximum fluctuation of speed. The ratio of the maximumfluctuation of speed to the mean speed is called the co -efficient of fluctuation ofspeed.


►Questions :-

1. Explain the turning moment diagram for two stroke & four stroke cycle ICengine.

2. What is the function of flywheel ? How does it differ from governor ?3. Explain the following term

(i) Coefficient of fluctuation of energy.(ii) Coefficient of fluctuation of speed.

4. Difference between flywheel & governor.

Problem :- 1A vertical double acting steam engine develops 75 kw at 250 rpm. The maximumfluctuation of energy is 30 % of work done. The maximum & minimum speeds are not tovary more than 1 % on either side of mean speed. Find the mass of the flywheel required,if the radius of gyration is 0.6 m.

Problem :- 2The turning moment diagram for a four cylinder petrol engine is draw to the followingscale turning moment 1 Cm = 20 Nm & Crank displacement 1 Cm = 15 o. The fluctuationof energy for crank position for minim um & maximum speeds corresponding to an axisof 4 Cm2 of turning moment diagram. A flywheel of dish type is to be fitted to the engine& required to have total fluctuation of speed equal to 1.5 % of mean speed of 1200 rpm.If outside diameter of flywheel i s 50 Cm, Determine the weight of flywheel required.

Problem :- 3In punching operation 1200 NM energy is required to punch a hole and the time ofpunching operation is 1.5 sec. Flywheel is rotated at 300 rpm. before punching the massof flywheel is 150 kg. & 307 kw power is required to operate the punching machine.Calculate the immediate speed of flywheel after punching operation. How manypunching operation will be in one minute.

Problem :- 4In a turning moment diagram of multi cylinder engine , the area above below the meantorque line taken in order are 5.81 , 3.23 , 3.87 , 5.16 , 1.94 , 3.87 ,2.58 & 1.94 Cm 2

respectively. The scale of turning moment diagram are,Vertical scale :- Turning moment 1 Cm = 7000 N.m.Horizontal scale :- Crank angle 1 Cm = 60 o.

Calculate the maximum fluctuation of energy.




Learning and understanding Types of governor , function of governor Centrifugal governor. Terminology governor

►Introduction:The function of a governor is to regulate the mean speed of an engine,

when there are variations in the load e.g. when the load on an engine increases, itsspeed decreases, therefore it becomes necessary to increase the supply of workin g fluid.On the other hand, when the load on the engine decrease, its speed increases andhence less working fluid is required.The Governor automatically controls the supply of working fluid to the enginewith the varying load condition and keeps the mean speed within certain limits.

A l i t t le consid eratio n will sho w that wh en the lo a d in crease, theconfiguration of the governor changes and a valve is moved to increase the supplyof working fluid: when the load decreases, the en gine speed increases and the governordecreases the supply of working fluid.

►Types of governor :The governor may, broadly be classified as below;

Centrifugal governorsInertia governors

The Centrifugal governors may further be classified as follows:

Centrifugal Governors

Pendulum type loaded type

Watt governor dead weight governors spring con governors

Porter governor Proell governor

Hornell governor hattung governor Wilson-hart Nell Pickering governor


THEORY OF MACHINE 69►Terms used in governors :

Height of governorI"It is the vertical distance from the centre of the ba ll to point where the axesof the arms. Intersect on the spindle axis. It is usually denoted by H

Equilibrium speed:It is the speed at which the governor balls , arms etc. are in completeequilibrium and the sleeve do not tend to move upwards or downw ards.

Mean equilibrium speed:It is the speed at the main position of the balls or the sleeve .

Maximum & minimum equilibrium speed:The speed at the maximum &- minimum radius of rotation of the balls, Without tendin gto move either way are known as maximum & minimum equilibrium speedSleeve liftIt is th e v ert i cal di stance which t he sl eeve t ravels due t o ch an ge in

equilibrium speed

►Sensitiveness of governors :Consider two governors A & B running at the same speed when this

speed increases o f decreases by a certain amount, the lift of the sleeve ofgovernor. A is greater than the lift of the sleeve of governor B. It is then saidthat the governor A is more sensitive than the governor 13.

In general , the greater the lift of the sleeve correspo nding to a givenfractional change in speed, the greater is the sensitiveness of the governor. Itm a y a l s o b e s t a t ed i n an o t h e r w a y t h at f o r a gi v en l i f t o f t h e s l ee v e , t h esensitiveness ma y be quite satisfactor y when the governor is considered as anindependent mechanismLet. N = m i n i m u m eq u i l i b r i u m s p eed

N = max i mu m equ i l i b ri um s p eed

N = mean equilibrium speed = NI + N1 / 2

Sensitiveness of the governor = N2 - N 1 / N= 2 (N2 - N1) / N1+N2= 2 (W 2 - W1) / W 1+W2

►Stability of governors :A governor is said to be stable when for every speed within the working

range there is a definite configuration i.e. there is only one radius of rotationof' the governor balls at which the governor is in equilibrium, for a stablegovernor, if the equilibrium speed increases the radius of governor balls must alsoincrease.

THEORY OF MACHINE 70►Isochronous governors :

A governor is said to be isochronous when the equilibrium speed isconstant (i.e. ran ge of speed is zero) for all radii of rotation of the balls withinthe working range, neglecting friction. The isochronous is the stage of infini tesensitivity.For Isochronous range of speed should be zero N2 - N1 = 0 or N2 = N1

►Hunting :A governor is said to be hunt if the speed of' the engine fluctuates

continuously above and below the mean speed. This is causes by a twosensitive governor which changes the fuel supply by a lar ge amount when asmall change in the speed of rotation takes place for example, when theload on the engine increases. the en gine speed decreases an d, if the governoris very sensitive, thegovernor sleev e immediately falls to its lowest position. This will resultin the opening of the control valve wide which will supply the fuel to theengine in excess of its requirement so that the engine speed rapidly increasesagain and the governor sleeve rises to it s highest position.

Such a governor may admit either the maximum or the minimumamount of the fuel. The effect of this will be to cause wide fluctuations inthe engine speed or in other words the engine will hunt.

►Effort and power of a governor :The effort of a governor is the mean force exerted at the sleeve for a

given percenta ge change of speed. It may be noted that when the governoris running steadily. There is no force at the sleeve. But when the speedchanges, there is a resistance at the sleeve w hich opposes its motion. It isassumed that this resistance which is equal to the effort, varies uniformlyfrom a maximum value to zero while the governor moves into its new position ofequilibrium.

The Power of a governor is the work done at the sleeve f or a givenpercentage change of speed. It is the product of the mean value of the effort andthe distance through which the sleeve moves, Mathematical,

Power = mean effort x lift of sleeve►Questions :-

1. Explain the following term.(i) Sensitivity of governor.(ii) Stability of governor.(iii) Hunting of governor.(iv) Governor power.

2. Explain governor effort.3. Give the classification of governors.4. Different between flywheel & governor.5. Explain the working principle of simple watt governor.6. What is the function of governor ? How is differ from flywheel ?




Learning and understanding Types of Balancing

Needs of balancing

Static and dynamic balancing .

►Introduction:T h e h i g h s p e e d o f e n g i n e s a n d o t h e r m a c h i n e s i s a c o m m o n

phenomenon now days. It is therefore very ess ential that all the rotating and reciprocatingpart should be completely balanced, the dynamic forces are setup. These forces not onlyincrease the loads on bearin gs and stresses in the v a r i o u s m em b e rs b u t a l s op r o d u c e u n p l e as an t n o i s e an d ev e r d an g e ro u s vibrations. We shall discussthe balancing forces caused by rotatianci g masses, in order of minimize pressureon the main bearings when an engine is running.

The center of gravity of a solid mass having uniform cross section throughout its length will be at it center. But in case of solid mass which is not havinguniform cross section, the center of gravity will be away from the center of thebod y. When such bod y not having uniform cross section is made to revolve,then it starts vibrating. So to a void vibrations of body such body is balanced byproviding counter weight on a unbalanced body or party.

►Concept :All rotating masses or machine, which is running at very high speeds,

requires the balancing of rotating masses. At very high speed the centrifugalforce or inertia force will be higher. If this high speed machines are usedwithout balancing it, then there will be possibility of producing vibrations willgive adverse effect to the performance of the machines.

►Need:In a running machine, there are two types of forces acting, static and

inertia forces, the static forces due to workin g fluid and weight of machine it self.The inertia forces due to rotation of various component of engine. Such forces, ifnot in equilibrium may form resultant force and couple. The purpose of balancingis to eliminate such resultant force and couple between machine frame andfoundation. Thus the purpose of balancing is to avoid vibration of machine bybalancing the inertia resultant force and couple.


►Types of balancing

1 Static BalancingThe balancin g of static parts of a mach ine by stat ic load is known as

static balancing. Example- balancing the frame of machine.

2 Dynamic BalancingThe balancing of unbalanced rotating mass by using rota ting mass of

appropriate weight is called dynamic balancing

Balancing of several masses rotating in the same plane


►B a l a n c i n g o f s e v e r a l m a s s e s r o t a t i n g i n t h e s a m e plane:Consider any number of masses (say four) of magnitude m l, m2, m3 and m4 atdistance of the rotatin g shaft. Let r l , r2 , r3 & r4 from the axis of the r o t a t i n gs h a f t . L e t θ 1 θ 2 , θ 3 a n d θ 4 b e t h e a n g l e s o f t h e s e m a s s e s a n dperpendicular to the plane of paper with a constant angular velocity of ( ω) rad / sec.The magnitude & position of balancing mass may be found out analytically or graphically asdiscuss below.There are two methods for balancing the several masses rotating in the same plane.

(a) Analyti cal Method1. First of all finds out the centriftigal force of each rotating mass .2. Resolve the centrifugal forces horizontally and vertically and find their

sums i.e. EH and EV.EH = m1•r1COS θ 1 + m2•r2COS θ 2 + m3•r3COS θ 3 + m4•r4COS θ 4EV = m1•r1SINθ 1 + m2•r2SIN θ 2 + m3•r3SIN θ 3 + m4•r4SIN θ 4

3. Magnitude of the resultant centrifugal force

R = √(EH)2 + (EV)2

4. If θ w i l l be the angle which the resultant force makes withhorizontal then,

Tanθ = EV / EV4. The balancing force is then equal to the resultant force but in

opposite direction.5. Now to find out weight of balancing mass

M x r = RWhere m= mass of balancing

R = radius of Rotation(b) Graphical Method1. Draw the space diagram with the position of the several masses as shown in

Figure.2. Find out the centrifugal force produce b y each mass i .e. m1•r1 m2•r2

m3•r3 & m4•r43. Take suitable scale and draw the vector diagram with obtained

centrifugal force such that ab = m1•r1 and parallel to OM1, be = m2•r2 andparallel to OM2. Similarly others cd and de can be drawn.

4. Now closing the side ae represents resultant force in magnitude and direction asshown in figure.

5. The balancing force is then equal to the resultan t force but in oppositedirection. R = ae x scale

6. Now, to find out weight of balancing massM x r = R


►Questions : -

1) Distinguish between2) Static balancing and dynamic balancing3) Primary and secondary force.4) Explain the balancing of several masses revolving in the same5) Plan by analytical method.6) How reciprocating mass is balanced?7) Explain what is balancing? State the its effects of unbalanced mass on

machine.8) Prove that if a body is in dynamic balance, it will be in static balance, but

the reveres is not true.9) Why balancing of rotating & reciprocating parts of engine is necessary?



Learning and understandingTypes of vibrationMotions of vibrationTerminology of vbration

►Intoduci ion :When elastic bodies such as a spring, a beam and a shaft are displaced from the

equilibrium position by the application of external forces, and then released, they executea vibratory motion. This is due to the reason that. when a bod y is displaced. the internalforces in the form of elastic or strain ener gy are present in the bode at release these forcesbring the body to its original position. When the bod y reaches the equilibrium position,the whole of' the elastic or strain ener gy is converted into kinetic energy due to which thebody continues to move in the opposite direction. The whole of the kin etic energy is againconverted into strain energy due to which the body again returns to the equilibriumposition. In this way the vibratory motion is repeated indefinitely.

►Terms used in vibratory motion :The following terms are commonly used in connection with the vibratory motions:

Period of vibration or Time Period:It is the time interval after which the motion is repeated itself. Theperiod of vibration is usually expressed in seconds.Cycle :It is the motion completed during one time period.Frequency:It is the number of cycles described in one second in s .i. units the frequency isexpressed in hertz which is equal to one cycle per second .

►Types vibratory motion: The following types of vibratory motion areimportant from the subject point of view.

1. Free or Natural Vibration:When no ex ternal force acts on the bod y, after giving an initial

displacement. then the body is said to be under free or natural vibration. Thefrequency of the free vibration is called free. or natural frequency.


2 Forced Vibrations:When the body vibrates under the influence of external force, then the

body is said to be under forced v ibration. The external force applied to the body is aperiodic disturb ing force created by unbalance. The vibrations have the samefrequency as the applied forces.


3 Damped Vibrations:When there is a reduction in amplitude over every cy cle of'

vibrations, the motion is said to be damped vibration. This is due to the factthat a certain amount of energy possessed by the vibrating system is alwaysdissipated. in over coming frictional resistance to the motion.

►Types of free vibration :The following three types of free vibrations are important from the

subject point of view.

1. Longitudinal Vibration :When the particles of the shaft or disc move parallel to the axis of the

shaft as shown in fi gure, then the vibrations ar e known as longitudinalvibrations. In this case the shaft is elongated and shortened alternatively andthus the tensile and compressive stresses are induced alternatively in the shaft

2 Transverse Vibrations:W h e n t h e p a r t i c l e s o f t h e s h a f t o r d i s c m o v e a p p r o x i m a t e l y

perpendicular to the axis of the shaft, as shown in fi gure then the vibrationsare known as transverse vibrations. In this case, the shaft is straight andbent alternatively and bending stresses are induced in the shaft.


3 Torsion Vibrations:When the particles of the shaft or disc move in circle about the axis of

t h e sh af t as sh o wn i n f i gu re th en th e vi b rat io ns are k n o wn as t o rsi n alvibrations, in this case, the shaft is twisted & untwisted alternately andthe torsion shear stresses are induced in the shaft.

►Critical or whirling speed of a shaft :In actual practice, a rotating shaft carries different mountings and

accessories in the form of gears or pulleys etc. when the gears or pulleys are put on theshaft, the cent er of gravity of the pulley or gear does not coincide with thecenter line of the bearings or with the axis of the shaft, when the shaft isstationary. This means that the center of gravity of the pulley or gear is at acertain distance from the axis of ro tation and due to this the shaft is subjected tocentrifugal force. This force will bent the shaft which will further increase thedistance of-center of gravity of the pulley or gear from the axis of rotation. Thiscorrespondingly increases the value of centrifugal force, which furtherincrease t h e d i s t an ce o f cen t er o f g rav i t y f ro m t h e ax is o f rot at i o n . Th i seffe ct i s cumulative and ultimately the shaft fails. This bending of shaft not onlydepends upon the value o f eccentricity but also depends upon the speed atwhich the shaft rotates. The speed, at which the shaft runs so that the additionaldeflection of the shaft from the axis of rotation becomes infinite, is known ascritical or whirling.

►Causes of vibrations:

-- Loose fitting-- Incorrect alignment-- Vibration waves-- Lack of isolation-- Lack of compact soil-- Insufficient balance (unbalancing )

►Remedies to reduce vibration:

-- Do live loose fittings-- Make correct alignment.-- Provide sufficient balancing & Reduce vibration-- Use isolation pads-- Use of compact soil


►Questions :-

1. Define the following term,(i) Free vibration(ii) Damped vibration(iii) Forced vibration.(iv) Degree of freedom(v) Critical speed(vi) Resonance(vii) Damping

2. State the cases of vibration and method for reducing vibration.3. Explain the model of vibrating system and explain its each major elements with a

brief description.